Final Exam_Solution Key

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STUDENT NAME 1 ChBE 3225 SEPARATION PROCESS PRINCIPLES Spring 2018 FINAL EXAM Wednesday, May 2, 2018 Open Textbook + 1 double-sided page of Notes This exam booklet consists of 3 problems, each worth the indicated points. The total number of points is 100. Please make sure that you write your name on the top of this page and sign at the bottom. Please make sure to read the entire problem statement carefully, including the numerical data and other information provided at the end of the problem. Please submit your answers in the underlined spaces below each question in the units required. Use the blank spaces provided in this booklet to work out the problems. Read and Sign: On my honor, I pledge that I have neither given nor received inappropriate aid on this exam. Signature

STUDENT NAME 2 Problem 1 (30 points total): Component A is to be separated from a binary mixture with component B, using liquid extraction with a pure solvent C. The required products are an extract containing 70 wt% A and a raffinate containing 3 wt% A, both expressed on a solvent-free basis . An existing mixer-settler cascade of 3 stages is available, and it is desired to evaluate its capabilities for this application. The stage efficiency is assumed ~100%. A. (5 points): Using the data below, draw the ternary phase diagram and tie lines. Use the graph sheet provided (two copies are provided for convenience). B. (8 points): Label the locations of the actual (solvent-included) extract and raffinate streams. C. (10 points): Feeds with high concentrations of A could potentially be handled with fewer than 3 stages, whereas very dilute feeds (i.e., low concentrations of A) may require more than 3 stages. Hence, it is desired to estimate the composition of the most dilute feed that can be handled by the 3-stage cascade while maintaining the required product specifications. Using a trial-and-error method, estimate the composition of this feed. Locate it on the phase diagram, and enter the result here: _______ wt% A, _______ wt% B, and 0 wt% C. D. (7 points): How much feed can be handled per unit (say, 100 kg/hr) of solvent? Answer: _______ kg/hr. Data for equilibrium curve: Extract phase mass fractions Raffinate phase mass fractions A B C A B C 0.05 0.00 0.95 0.05 0.95 0.00 0.10 0.01 0.89 0.10 0.90 0.00 0.15 0.01 0.84 0.15 0.84 0.01 0.20 0.02 0.78 0.20 0.78 0.02 0.25 0.03 0.72 0.25 0.71 0.04 0.30 0.04 0.66 0.30 0.63 0.07 0.35 0.07 0.58 0.35 0.50 0.15 0.40 0.13 0.47 0.40 0.26 0.34 Data for tie lines: Extract phase weight fraction of A Raffinate phase weight fraction of A 0.40 0.31 0.22 0.15 0.13 0.09 0.08 0.07 0.04 0.04

STUDENT NAME 3

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STUDENT NAME 4 A. The right-triangle diagram is shown below. B. Solvent-free extract is labeled P. The actual extract is the combination of P and pure solvent S, and must lie on the equilibrium curve. Hence we locate the actual extract E. Similarly, we locate the raffinate R. Note that due to the shape of the equilibrium curve, the solvent-free and solvent-included raffinate are nearly the same (i.e. there is very little solvent in the raffinate). C. Since the stage efficiency is nearly 100%, the number of equilibrium stages = the number of actual stages = 3. However, the feed composition is unknown. The feed is solvent- free, so it must be located somewhere between P and R. Trial-and-error method is applied to determine F: 1. Locate an arbitrary F between P and R. 2. Elongate lines FE and RS to get the operating point. 3. Using the operating point and the extract E, find the number of stages to reach R by the graphical method. 4. If the number of stages is 3  Done! 5. If not  Re-guess an F and repeat the above steps. The result is shown on the left. The feed composition is approximately 58% A and 42% B. Alternatively, we can guess a mixture point M = F+S, that must also lie on the line ER. Extrapolating the line SM to the vertical axis gives the corresponding feed. The rest of the procedure is the same as above. D. The solvent to feed ratio (S/F) can be easily found by measuring the length ratio of lines FM and SM. This ratio is found to be 0.56. For a solvent flow of 100 kg/hr, the cascade can handle a feed flow of 100/0.56 = 179 kg/hr of feed.

STUDENT NAME 5 Problem 2 (35 points total): Furfural is an important biomass-derived chemical. A dilute (3 mol% furfural / 97 mol% water) stream of 1 m 3 /min is produced by a catalytic process. Since furfural has a much higher boiling point than water and also forms an azeotrope with water, direct distillation to produce 99% pure furfural is not feasible. It is decided to use a combined adsorption and membrane process. The cyclic adsorption process uses a furfural-selective, highly hydrophobic molecular sieve adsorbent packed bed to concentrate the furfural in the adsorbent pellets at 25°C, whereas the exit (raffinate) stream - which is almost pure water with a very small amount of unadsorbed furfural - is recycled upstream. The adsorption step is stopped at the appropriate breakthrough time. Then the adsorbed furfural is recovered by purging the bed with hot air (at 100°C), and the liquid furfural product is recovered from the air using a total condenser (i.e., the air exiting the condenser is almost completely dry). The air purge removes all the furfural adsorbed inside the adsorbent pellets. However, it also removes the water-rich liquid trapped in the spaces between the adsorbent pellets (i.e., in the void fraction  b ). As a result, the purity of the recovered product from the adsorption column is only 94 mol% furfural. To remove the remaining water, the adsorption product is sent to a highly hydrophilic molecular sieve pervaporation membrane. The furfural-rich stream from the adsorption bed enters at a rate of 50 kmol/hr. The membrane operates at 50°C with a feed-side pressure of 101 kPa and a permeate-side pressure of 0.1 kPa. It generates a retentate stream of the required 99% pure furfural. A . (15 points): Assuming that the adsorbent bed is practically empty at the beginning of the adsorption cycle, the breakthrough time of the adsorption step (defined as the time required for the exit stream concentration of furfural to reach 0.03 mol%) is _________ minutes. B. (8 points): The water flux is ____________ kmol/m 2 /hr and the furfural flux is ____________ kmol/m 2 /hr. C. (4 points): The furfural content of the permeate stream is ___________ mol%. D. (4 points): The production rate of the 99% furfural stream is ___________ kmol/hr. E. (4 points): The required pervaporation membrane area is __________ m 2 . Adsorption Data: Furfural obeys a linear adsorption isotherm q *= 8 c* in the molecular sieve. Water is not significantly adsorbed by the hydrophobic molecular sieve. The bed has a cross-sectional area of 0.5 m 2 , 8 m length, void fraction  b = 0.5. The overall mass transfer coefficient ( k ) is 0.5 min -1 . Membrane Data: Perfect mixing can be assumed on both sides of the membrane. Assume that the permeate-side pressure is negligible during flux calculations. Permeances (kmol/m 2 /hr/kPa) through membrane: furfural = 1x10 -5 and water = 2x10 -3 At 94 mol% furfural concentration and 50°C, the activity coefficients are 1.2 (furfural) and 2.5 (water). At 50°C the saturation vapor pressures are 1.6 kPa (furfural) and 13 kPa (water).

STUDENT NAME 6 Other : Approximation of the error function: erf( x ) ~ sin ( x 2/3 ) with x obtained in radians and not degrees. A. The breakthrough time is reached when C/C F at the bed exit reaches 0.03/3 = 0.01. The breakthrough time can be estimated from the Klinkenberg equation with C/C F = 0.01. Hence we get: 0.01 = 0.5 [1+erf (X)] where X = √  - √  +1/8√  +1/8√  Using the erf approximation given, we get X = -1.6. Superficial velocity of liquid in the bed = 1 m 3 /min / 0.5 m 2 = 2 m/min. Actual (interstitial) velocity u is therefore 2/  b =4 m/min. At the bed exit, z = 8 m. Hence  = ( kKz/u ). (1-  b )/  b = 0.5x8x8/4 = 8. So we get X = -1.6 = √  - √  +1/8√  +1/8√    = 1.138 = k. ( t - z/u ) = 0.5 x ( t – 8/4)  t = 4.28 min B. Since the perfect mixing assumption is valid, the liquid-side composition in contact with the membrane is uniformly the retentate value x R = 0.99. At this composition, the furfural activity coefficient is given as 1.1. The flux of any permeant is given by: ( ) sat i Mi i i i i P N P x P y P    Since the permeate side pressure is negligible, we can calculate the fluxes: N furfural = 1x10 -5 (1.1x0.99x1.6) = 1.74x10 -5 kmol/m 2 /hr and N water = 2x10 -3 (4.5x0.01x13) = 1.17x10 -3 kmol/m 2 /hr. C. On the permeate side, the ratio of furfural and water mole fractions will be the ratio of their fluxes through the membranes, i.e. y furfural /y water = 0.0149 => y water = 0.9853, y furfural = 0.0147. D. Feed rate into the membrane = n F = 50 kmol/hr with x furfural = 0.94. Retentate has x = 0.99 and permeate has y = 0.0147. Therefore, mass balances can be used to calculate n R . n P + n R = 50 0.0147n P + 0.99n R = 0.94x50  This gives n R = 47.44 kmol/hr (furfural product) and n P = 2.56 kmol/hr. E. Since n P = 2.56 kmol/hr and y water = 0.9853, the water flow rate through the membrane = 2.56x0.9853 = 2.522 kmol/hr. Since the water flux is 1.17x10 -3 kmol/m 2 /hr, the required area = 2.522/1.17x10 -3 = 2156 m 2 .

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STUDENT NAME 7

STUDENT NAME 8 Problem 3 (35 points total): A saltwater stream (3.5 wt% NaCl) at 25°C and 1 atm is to be purified to produce 10 m 3 /hr of drinking water (0.05 wt% NaCl or lower). Two options are available: #1: An evaporator operating at 80°C to generate water vapor from the saltwater. The vapor is then condensed to produce liquid water. #2: A reverse osmosis membrane operating at 25°C with a feed pressure of 150 atm, permeate pressure of 1 atm, permeance of 5×10 -8 m 3 water/m 2 .s.atm, and powered by a high-pressure feed pump operating with an electrical efficiency (  ) of 60% (i.e., 0.6). The volumetric cut (   Q permeate /Q feed ) is 0.5. A. (4 points): The NaCl concentration in the saltwater is _______ mol/m 3 and that of water is ________ mol/m 3 . B. (4 points): Assuming the evaporator produces pure water vapor and a brine with 15 wt% NaCl, the volume flow rate of the brine leaving the evaporator is _________ m 3 /hr. C. (7 points): The heat energy requirement of the evaporator is _________ kW (1 W = 1 J/s). D. (8 points): The flux through the RO membrane is ________ m 3 water/m 2 /hr. E. (4 points): The required membrane area is _________ m 2 . F. (8 points): The energy cost of the evaporator is $_________/ year and that of the RO membrane is $_________/ year. Data and Assumptions: Molecular weights: NaCl = 58.5, water = 18. All solutions can be assumed to have density ~ 1000 kg/m 3 . 1 atm = 14.7 psia = 1.013×10 5 Pa. Enthalpy data: At 25°C, enthalpy of 3.5 wt% NaCl aqueous solution: 106 kJ/kg At 80°C, enthalpy of 15 wt% NaCl aqueous solution: 340 kJ/kg At 80°C, enthalpy of saturated water vapor: 2643 kJ/kg Evaporator: Heating cost (based upon natural gas price and heat efficiency of the evaporator) is $4x10 -6 /kJ. RO Membrane: Assume concentration polarization effects to be negligible. Osmotic pressure can be calculated using the same equation as used for seawater. The electrical power requirement for the high-pressure feed pump is given by: P e (kW) = 2.78×10 -7 . Q . ( P f - P i )/  , where Q = volume flow rate in m 3 /hr, P f and P i = pressures (in Pa) of liquid exiting and entering the pump, and  = electrical efficiency. Electricity cost is $1.94x10 -5 /kJ.

STUDENT NAME 9 A. Feed salt concentration = 3.5 wt% NaCl = (0.035/58.5 mol/g saltwater)x(1000 g/kg saltwater). (1000 kg/m 3 ) = 598 mol/m 3 . Feed water concentration = 96.5 wt% H 2 O = (0.965/18 mol/g saltwater). (1000 g/kg saltwater). (1000 kg/m 3 ) = 53611 mol/m 3 . B. The evaporator must produce 10 m 3 /hr water vapor = 10,000 kg/hr water (which has no NaCl) = n W . Brine flow rate = n B and feed flow rate = n F . Overall mass balance: n F = n W + n B Salt balance: 0.035n F = 0.15n B Hence we get n B = 3044 kg/hr  volume flow rate of 3.044 m 3 /hr. C. Using the given enthalpy data, energy balance on evaporator: 106n F + Q = 2643n W + 340n B  Q = 26082296 kJ/hr = 7245 kW. D. We have  = 1.12x298x2x(598 - 598x0.05/3.5)/(1000 L/m 3 ) = 393.5 psia =26.8 atm Drinking water flux J = 5x10 -8 .(149-26.8) = 6.11x10 -6 m 3 /m 2 /s. E. Since the production rate is Q permeate = 10 m 3 /hr, the required area is A M = (10/3600)/6.11x10 -6 = 455 m 2 . F. RO membrane: Since the volumetric cut is 0.5, the feed volume flow rate is 20 m 3 /hr. Electrical power 7 5 2.78 10 (10/ 0.5) (149) 1.013 10 / 0.6 140 P kW         Hence, the electricity cost will be 140 kJ/s x $1.94x10 -5 /kJ = $2.722x10 -3 /s = $86,000/yr Evaporator: the heating energy cost will be 7245 kJ/s. $4x10 -6 /kJ = $2.898x10 -2 /s = $914,000/yr

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